Question

$\int\limits_{0}^{\pi / 2} \frac{d x}{4+5 \sin x}=$

• A. $\frac{1}{2} \log 3$
• B. $\frac{1}{3} \log 2$
• C. $2 \log 3$
• D. $\frac{1}{2} \log \frac{3}{2}$

Answer 1

Answer: (B)

We have, $\int\limits_{0}^{\pi / 2} \frac{d x}{4+5 \sin x}$
$=\int\limits_{0}^{\pi / 2} \frac{d x}{4+5\left(\frac{2 \tan \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}}\right)}$
$=\int\limits_{0}^{\pi 2} \frac{\left(1+\tan ^{2} \frac{x}{2}\right)}{4+4 \tan ^{2} \frac{x}{2}+10 \tan \frac{x}{2}} d x$
$=\frac{1}{4} \int\limits_{0}^{\pi 2} \frac{\sec \frac{2 x}{2}}{\tan ^{2} \frac{x}{2}+\frac{5}{2} \tan \frac{x}{2}+1} d x$
let $\tan \frac{x}{2}=t$
$\Rightarrow \sec ^{2} \frac{x}{2}\left(\frac{1}{2}\right) d x=d t$
$\Rightarrow \sec ^{2} \frac{x}{2} d x=2 d t$ at $x=0, t=0$ and
$x=\frac{\pi}{2}, t=1=\frac{2}{4} \int\limits_{0}^{1}-\frac{1}{t^{2}+\frac{5}{2} t+1} d t$
$=\frac{1}{2} \int\limits_{0}^{1} \frac{d t}{t^{2}+\frac{5}{2} t+1+\frac{25}{16}-\frac{25}{16}}$
$=\frac{1}{2} \int\limits_{0}^{1} \frac{d t}{\left(t+\frac{5}{4}\right)^{2}-\frac{9}{16}}=\frac{1}{2} \int\limits_{0}^{1} \frac{d t}{\left(t+\frac{5}{4}\right)^{2}-\left(\frac{3}{4}\right)^{2}}$
$=\frac{1}{3}\left[\log \left|\frac{t+\frac{1}{2}}{t+2}\right|\right]_{0}^{1}$
$=\frac{1}{3}\left[\log \left(\frac{3 / 2}{3}\right)-\log \left(\frac{1 / 2}{2}\right)\right]$
$=\frac{1}{3}\left[\log \left(\frac{1}{2}\right)-\log \left(\frac{1}{4}\right)\right]$
$=\frac{1}{3}\left[\log \left(\frac{\frac{1}{2} 4}{1}\right)\right]$
$=\frac{1}{3} \log 2$